Pre algebra work
Introduction :
Math pre algebra work is a common name for a course in middle school mathematics. pre algebra work includes several broad subjects are Review of natural number arithmetic, new types of numbers such as integers, fractions, decimals and negative numbers
Pre algebra work often includes some basic subjects from geometry, mostly the kinds that further understanding of algebra and show how it is used, such as area, volume, and perimeter. (Source.Wikipedia)
pre algebra Special products work:
Let us consider three special product which occur very frequently in algebra They are :
(i) (a+b)2 =a2+2ab+b2
(ii) (a-b)2 =a2 -2ab +b2
(iii) (a+b)(a+b)=a2+b2
Examples:
(i) (4+ 3)2 = 42 + (2 *4 *3) + 32 or 49 = 49 or 72
(ii) (6− 3)2 = 62 − (2 *6* 3) + 32 or9= or 32
(iii) (3+ 3)(3− 3) = 32 − 32 or0= 0
Example 1:
Evaluate (a) (9x + 3y) 2 (b) (2x2 + 5y2) (2x2 − 5y2 )
(a) (9x + 3y)2 = (9x)2 + 2 ( 9x 3y )+ (3y)2
=81x2 +54xy + 9y2
(b) (2x2 + 5y2 )(2x2 − 5y2 ) = (2x2 )2 − (5y2 )2
=4x4 −25y4
pre algebra Common factors of Monomials work:
Example 1:
Factorize 49m2 - 36n2
Solution: We have 492- 36n2 = (7m)2- (6n)2
= (7m+6n) (7m-6n)
Example 2:
Factorize 100a2-(x-y)2
Solution:
We have
= [10a+(x-y)] [10a-(x-y)]
= (10a+x-y) (10a-x+y)
Example 3:
Factorize7x4 -7y4
Solution: We have
7( x4- y4)=7 [(x2)2 - (y2)2 ]
=7[(x2 + y2) (x2 - y2)]
=7[(x2 + y2) (x - y) (x + y)]
Q [x2 - y2 = (x - y) (x + y)]
Here, we note that the given binomial is complete by factorized since none of the factors can further be expressed as product of two expressions. We therefore, say that an expression is said to be completely factorized if none of the factors can be factorized, expressed as a product of the two expressions with lowest exponents of terms.
Example 3:
Find the greatest common factors of 2x5 and 14x2 and then factorize 2x5+14x2
Solution:
Here, the largest numbers that divides the coefficients of 2x5 is 5 and
14x2 is 2. The highest power of x that is a factor of x5 and x2 . The greatest common factor of 2x5 and 14x2 is therefore 2x2
Thus, 2x5 + 14x2 =( 2x2.x3) +( 2x2.7)
= 2x2 (x3 + 7)
So, given expression can be expressed as the product of the two factors one of which is the greatest common factor
Work on problem for pre algebra:
1. Factorize 256x2 – 9
Answer: (16x+3) (16x-3)
2. Find the value of x if 6x = 352 -92
Answer: 190.66667
Math pre algebra work is a common name for a course in middle school mathematics. pre algebra work includes several broad subjects are Review of natural number arithmetic, new types of numbers such as integers, fractions, decimals and negative numbers
Pre algebra work often includes some basic subjects from geometry, mostly the kinds that further understanding of algebra and show how it is used, such as area, volume, and perimeter. (Source.Wikipedia)
pre algebra Special products work:
Let us consider three special product which occur very frequently in algebra They are :
(i) (a+b)2 =a2+2ab+b2
(ii) (a-b)2 =a2 -2ab +b2
(iii) (a+b)(a+b)=a2+b2
Examples:
(i) (4+ 3)2 = 42 + (2 *4 *3) + 32 or 49 = 49 or 72
(ii) (6− 3)2 = 62 − (2 *6* 3) + 32 or9= or 32
(iii) (3+ 3)(3− 3) = 32 − 32 or0= 0
Example 1:
Evaluate (a) (9x + 3y) 2 (b) (2x2 + 5y2) (2x2 − 5y2 )
(a) (9x + 3y)2 = (9x)2 + 2 ( 9x 3y )+ (3y)2
=81x2 +54xy + 9y2
(b) (2x2 + 5y2 )(2x2 − 5y2 ) = (2x2 )2 − (5y2 )2
=4x4 −25y4
pre algebra Common factors of Monomials work:
Example 1:
Factorize 49m2 - 36n2
Solution: We have 492- 36n2 = (7m)2- (6n)2
= (7m+6n) (7m-6n)
Example 2:
Factorize 100a2-(x-y)2
Solution:
We have
= [10a+(x-y)] [10a-(x-y)]
= (10a+x-y) (10a-x+y)
Example 3:
Factorize7x4 -7y4
Solution: We have
7( x4- y4)=7 [(x2)2 - (y2)2 ]
=7[(x2 + y2) (x2 - y2)]
=7[(x2 + y2) (x - y) (x + y)]
Q [x2 - y2 = (x - y) (x + y)]
Here, we note that the given binomial is complete by factorized since none of the factors can further be expressed as product of two expressions. We therefore, say that an expression is said to be completely factorized if none of the factors can be factorized, expressed as a product of the two expressions with lowest exponents of terms.
Example 3:
Find the greatest common factors of 2x5 and 14x2 and then factorize 2x5+14x2
Solution:
Here, the largest numbers that divides the coefficients of 2x5 is 5 and
14x2 is 2. The highest power of x that is a factor of x5 and x2 . The greatest common factor of 2x5 and 14x2 is therefore 2x2
Thus, 2x5 + 14x2 =( 2x2.x3) +( 2x2.7)
= 2x2 (x3 + 7)
So, given expression can be expressed as the product of the two factors one of which is the greatest common factor
Work on problem for pre algebra:
1. Factorize 256x2 – 9
Answer: (16x+3) (16x-3)
2. Find the value of x if 6x = 352 -92
Answer: 190.66667